NeoPets Riddles (Lenny Conundrums) and Answers Here

When I first read the question, I knew it was going to involve calculus. Sigh...I think I will wait until next week to upgrade my trophy. I give.

okay, this may come off as irrevelent but they said egg not ball so would the whole spherical thing still work?

Found this:

http://www.1728.com/sphere.htm

Enjoy.

It says its a perfect sphere after it says its and egg.

oh whoops, missed that part ^^;; thanks/sorry

Aaaargh, I typed +h instead of -h in my Excel formula! All that calculus to calculate the spherical cap volume formula (I didn't find the Wolfram site) and I mistype a sign and submit the wrong answer.

Nice work, markr, but you really ought to make people do some of the numerical calculations themselves.

gahh im so terrible at math!

so would the answer be 388889?

stupid lenny conundrum!

I got the same answer as markr but several hours later due to other commitments. Thanks for the formulae, was using wikipedia's formula for volume of spherical cap initially and had the extra variable of the radius of the base of the cap. Quite straightforward using markr's method.

Several very kind people have already given the FORMULA on how to get the correct answer. YOU need to plug in the data from the program and come to the answer on your own.

No one is just going to right out and give it to you numerically. This site is not about just giving you the answer. Some of us are good at puzzles and logic and some of us are math type people.. but we work together to find the method on how to get the answer.

Unfortunately, some weeks the answer gets posted right out in the open, but that's not the way most of us prefer it.

Begging for the answer just makes you look like an ass wanting to do none of the work and take all the credit anyway. Either chip in and help do the work, and spend 10 minutes solving the problem with the SOLUTION already given. If you only want the answer without needing to do ANY work, just go somewhere else.

Lenny Conundrum
Hey guys. Ok we need to set things straight, we don't know what volume of egg is displacing the water, i assumed it was 72% but now i think this is wrong. One persons comment was that it is simply 100% of the volume of the water displaced to the volume of the egg, this could be true.

If not i have no idea how to figure out the volume of egg that is displacing the water so that it is possible to figure out the volume of the egg

Round 252: Volume of sphere, given volume of spherical cap

The submerged portion of the sphere is represented by a shape called a "spherical cap". For a sphere of radius R, the surface area A and volume V are given by:

. . .sphere with radius R:

. . . . .A = 4 (pi) R^2

. . . . .V = (4/3) (pi) R^3

By "truncating" the sphere (that is, by chopping off a portion of it), we are left with part of a sphere, with the missing portion represented by a flat surface, being a circle with radius r. The "height" of the whole sphere is obviously 2R (or the length of the diameter); the height of the truncated sphere is "h".

For a spherical cap with flat-surface radius r and height h, formed from a sphere with radius R, the surface area A and volume V are given by:

. . .spherical cap with radius R, cap radius r,
. . .and height h:

. . . . .A = 2 (pi) R h

. . . . .V = (1/6) (pi) (3 r^2 + h^2) h

The relationship between the radius R of the sphere and the radius r of the flat surface of (the circle formed by) the spherical cap is given by:

. . . . .R = (r^2 + h^2) / (2h)

We are given that only 72% of the egg's total surface is wetted when allowed to float in water. This means that the surface area of the spherical cap is 0.72 of the surface area of the entire sphere. That is:

. . . . .2 (pi) R h = 0.72 [4 (pi) R^2]

Solving, we get:

. . . . .2 (pi) R h = 4 (pi) (0.72) R^2

. . . . .R h = 2 (0.72) R^2

. . . . .h = 2 (0.72) R

. . . . .h = 1.44 R

Plugging this into the relationship between the circle's radius r and the sphere's radius R, we get:

. . . . .R = (r^2 + h^2) / (2h)

. . . . .R = (r^2 + 2.0736 R^2) / (2.88 R)

. . . . .2.88 R^2 = r^2 + 2.0736 R^2

. . . . .(2.88 - 2.0736) R^2 = r^2

. . . . .0.8064 R^2 = r^2

. . . . .sqrt[0.8064] R = r

We are given that the submerged volume (that is, the volume of the spherical cap) is 280,000 cc's. Then:

. . . . .V = (1/6) (pi) (3 r^2 + h^2) h

. . . . .280,000 = (1/6) (pi) (3 (0.8064 R^2) + (2.0736 R^2)) (1.44 R)

. . . . .280,000 = (1/6) (pi) (2.4192 R^2 + 2.0736 R^2) (1.44 R)

. . . . .280,000 = (1/6) (pi) (4.4928 R^2) (1.44 R)

. . . . .280,000 = (1/6) (pi) (4.4928) (1.44) (R^3)

. . . . .280,000 = (1/6) (pi) (6.469632) (R^3)

. . . . .[(280,000) (6)] / [(pi) (6.469632)] = R^3

. . . . .1,680,000 / [6.469 632 (pi)] = R^3

Since the formula for the volume of the sphere involves R^3, we do not need to solve for the value of R itself. Instead, we can proceed directly to finding the volume:

. . . . .V = (4/3) (pi) R^3

. . . . .V = [(4/3) (pi) (1,680,000)] / [6.469 632 (pi)]

. . . . .V = [(4) (1,680,000)] / [(3) (6.469 632)]

. . . . .V = [ 6 720 000 ] / [ 19.408 896 ]

...and so forth.

(Hint: Plug the above into your calculator. Then round the value to the nearest whole number, which is not 388 889.)


I completely agree with your sentiment, and I think it is to be regretted that experience has consistently shown that many do indeed want to be "just handed the answer", complete with a guarantee that it's the right answer, and right away ASAP ASAP URGENT HELP HELP NOW!!!!!.

Unfortunately, the "somewhere else's" to which this sort goes are frequently tutoring forums, where they lie and say that the "Lenny" is really a make-or-break take-home math quiz ("I'll flunk for the entire year if you don't give me the right answer right now!!"). In fact, this "math homework" question has already been posted any number of times to Yahoo! Answers (with quite a few of these posts already having been deleted or rebuked); you will note, upon examination, that many of the remaining posts show that the text of the "Lenny" has been munged to disguise the actual origin, a tacit admission that the postings are illegitimate.

As a result of such ill behavior from Neopians, many other sites now customarily ban any user posting Neopets-related material, and are quite happy to see the Conundrums' solution posted, openly and as quickly as possible, in locations such as this one.

Sorry.

Eliz.

Aw, you are too kind. I would have quit at least 3 lines earlier.

Doing so would seem, upon a review of past evidence, to presume perhaps too much with respect to diligence and/or intelligence. Take a quick check back through this thread and discover how many times that week's solution has been posted along the lines of:


...and how many times the reply has been, "So what's the answer? Stop making it so hard and just tell me NOW!"

I'm currently awaiting complaints that what has been posted for Round 252 ("plug this number into your calculato, then hit the 'divide' key, and then plug the other number in; and round the result displayed on the calculator screen") is, once again, "too hard". And just wait until this discussion laps over onto a new "page". Hitting that "previous" link at the top of the page is apparently just way too confusing and complex!

Eliz.

Oh, you are so right, but I would personally love it if we were able to ignore such people. I do think that posting a full answer after the submissions are closed is a great idea, as Neopets often does a really bad job of showing how to get the answer and people certainly have trouble figuring it out.

During the contest, my preference for math problems would be to put out the relevant formulas and give an overview of how to approach the problem, but not include any calculations. I'll bet my math background biases me, however.

Anyway, I'm going to do my best to help the discussion move to a new page - I don't really want to share my neopoints with people that can't even navigate through web pages. (Not that I get any this time anyway, since I added when I should have subtracted in doing my own calculations.)

Hi
Darn, I submitted 388889 BEFORE i found this site.. xD

But thanks stapel for all the help! I was able to find the right answer.

I am not sure if I did it right. But I calculated that 72% of surface area is equal to 54% of the area. But I ended up with a different answer than all of you, not sure if this is mathematically legal though.

54% of the area of what? The only thing we have is a sphere and the only area I know of is the surface area of the sphere.

The general formula turns out to be quite simple:

V = SV / [SA^2 * (3 - 2 * SA)]

where:
SV = submerged volume
SA = submerged surface area as a fraction of total surface area

Lmao, U have to frickin calculate the are so that u can get it!! so what are u guys do over the 2 weeks that lenny is busy?

Thanks!
Thanks so much for working that out. I was looking for help with the Lenny Conundrum...it's been YEARS since I've worked with surface area and volume (no, I am not smarter than a fifth grader). Posting the way to get there helped to jog my memory a bit. So much better when I get to work through it, rather than just giving me the answer.